British astronomer Eddington liked to ride a bike. It is said that in order to show off his skill, he has even defined an "Eddington number", E -- that is, the maximum integer E such that it is for E days that one rides more than E miles. Eddington's own E was 87.
Now given everyday's distances that one rides for N days, you are supposed to find the corresponding E (≤N).
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤105), the days of continuous riding. Then N non-negative integers are given in the next line, being the riding distances of everyday.
Output Specification:
For each case, print in a line the Eddington number for these N days.
Sample Input:
10
6 7 6 9 3 10 8 2 7 8
Sample Output:
6
思路:从小到大排序,E从大到小考虑,只需要E个数中考虑最小的那个数,即v[n-E]是否大于E就行,不成立,E就减一;
程序:
#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
vector<int> v;
for(int i = 0; i < n; i++)
{
int a;
scanf("%d",&a);
v.push_back(a);
}
sort(v.begin(),v.end());
int E = v.size();
while(E)
{
if(v[n-E] <= E)
E--;
else
break;
}
printf("%d\n",E);
return 0;
}